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Bài 1:
a) \(\frac{2}{5}+\frac{1}{5}.\left(\frac{3}{4}\right)\)
= \(\frac{2}{5}+\frac{3}{20}\)
= \(\frac{11}{20}\)
b) \(\frac{5}{12}.\left(-\frac{3}{4}\right)\) + \(\frac{7}{12}.\left(-\frac{3}{4}\right)\)
= \(\left(\frac{5}{12}+\frac{7}{12}\right).\left(-\frac{3}{4}\right)\)
= 1.\(\left(-\frac{3}{4}\right)\)
= \(-\frac{3}{4}\)
Còn câu c) đang nghĩ.
Bài 2:
a) \(\frac{5}{7}+\frac{2}{7}\)x = 1
1.x = 1
x = 1 : 1
x = 1
Vậy x = 1.
b) 0,2 + | x - 1, 3 = 1, 5|
0,2 + x = 1, 5 + 1, 3
0,2 + x = 2, 8
x = 2, 8 - 0, 2
x = 2, 6
Vậy x = 2, 6.
c) 2x + 5 = 37
2x = 37 - 5
2x = 32
2x = 25
=> x = 5
Vậy x = 5.
d) 2x + 2x + 1 = 48
2x . 1 + 2x . 21 = 48
2x . ( 1 + 2) = 48
2x . 3 = 48
2x = 48 : 3
2x = 16
2x = 24
=> x = 4
Vậy x = 4.
Chúc bạn học tốt!
làm bước trung gian giùm mình luôn nhé
thanks trước những bạn làm giùm nhé
mình đang cần gấp lắm sáng mai là mình cần ai đang on làm giùm mình nhé
thanks
Bài làm :
a)\(=-\frac{3}{5}+\frac{28}{5}\times\frac{9}{14}=-\frac{3}{5}+\frac{18}{5}=3\)
b)\(=\frac{55}{126}+\frac{5}{42}+\frac{4}{9}=1\)
c)\(=-\frac{51}{13}-\frac{27}{13}=-6\)
d)\(=\frac{7}{3}-11\frac{1}{4}\times\frac{2}{15}=\frac{7}{3}-\frac{3}{2}=\frac{5}{6}\)
e)\(=1\times\frac{8}{3}\times0,25=\frac{2}{3}\)
a) \(\frac{2^{11}.9^2}{3^5.16^2}=\frac{2^{11}.3^4}{3^5.2^8}=\frac{2^3}{3}=\frac{8}{3}\)
Bài 2:
a) \(x:\left(\frac{2}{9}-\frac{1}{5}\right)=\frac{8}{16}\)
\(\Leftrightarrow x:\frac{1}{45}=\frac{1}{2}\)
\(\Leftrightarrow x=\frac{1}{2}:\frac{1}{45}=\frac{45}{2}\)
b) \(\left(2x-1\right).\left(2x+3\right)=0\)
\(\)\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\2x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=1\\2x=-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=-\frac{3}{2}\end{matrix}\right.\)
c) \(\frac{4-3x}{2x+5}=0\Leftrightarrow4-3x=0\)
\(\Leftrightarrow3x=4\Rightarrow x=\frac{4}{3}\)
d) \(\left(x-2\right).\left(x+\frac{2}{3}\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2>0\\x+\frac{3}{2}>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-2< 0\\x+\frac{3}{2}< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>2\\x>-\frac{3}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x< 2\\x< -\frac{3}{2}\end{matrix}\right.\end{matrix}\right.\)
Bài 2:
a) \(x:\left(\frac{2}{9}-\frac{1}{5}\right)=\frac{8}{16}\)
=> \(x:\frac{1}{45}=\frac{1}{2}\)
=> \(x=\frac{1}{2}.\frac{1}{45}\)
=> \(x=\frac{1}{90}\)
Vậy \(x=\frac{1}{90}.\)
b) \(\left(2x-1\right).\left(2x+3\right)=0\)
=> \(\left\{{}\begin{matrix}2x-1=0\\2x+3=0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}2x=0+1=1\\2x=0-3=-3\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x=1:2\\x=\left(-3\right):2\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=\frac{1}{2}\\x=-\frac{3}{2}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{1}{2};-\frac{3}{2}\right\}.\)
Mình chỉ làm được thế thôi nhé, mong bạn thông cảm.
Chúc bạn học tốt!
Câu 1:
\(A=3^2\cdot\frac{1}{243}\cdot8^2\cdot\frac{1}{33}\)
\(=\frac{3^2}{3^5}\cdot\frac{64}{33}\)
\(=\frac{1}{3^3}\cdot\frac{64}{33}=\frac{64}{33\cdot27}=\frac{64}{891}\)
\(B=\left(4\cdot2^5\right):\left(2^3\cdot\frac{1}{16}\right)\)
\(=2^2\cdot2^5:\left(\frac12\right)\)
\(=2^7\cdot2=2^8=256\)
\(C=\left(1\frac34\right)^3-\left(1\frac34\right)^3+\left(-1,031\right)^0\)
\(=\left(\frac74\right)^3-\left(\frac74\right)^3+1\)
=1
\(D=\frac{45^{10}\cdot5^{20}}{75^{15}}\)
\(=\frac{\left(3^2\cdot5\right)^{10}\cdot5^{20}}{\left(3\cdot5^2\right)^{15}}=\frac{3^{20}\cdot5^{10}\cdot5^{20}}{3^{15}\cdot5^{30}}\)
\(=\frac{3^{20}}{3^{15}}=3^5=243\)
\(E=\frac{3^{17}\cdot81^{11}}{27^{10}\cdot9^{15}}\)
\(=\frac{3^{17}\cdot\left(3^4\right)^{11}}{\left(3^3\right)^{10}\cdot\left(3^2\right)^{15}}\)
\(=\frac{3^{17}\cdot3^{44}}{3^{30}\cdot3^{30}}=\frac{3^{61}}{3^{60}}=3\)
Bài 2:
a: \(11^{x-1}=11^7\)
=>x-1=7
=>x=7+1=8
b: \(\left(x-4\right)^2=64\)
=>\(\left[\begin{array}{l}x-4=8\\ x-4=-8\end{array}\right.\Rightarrow\left[\begin{array}{l}x=8+4=12\\ x=-8+4=-4\end{array}\right.\)
c: \(5^{x+1}-5^{x}=100\cdot25^{29}\)
=>\(5^{x}\cdot5-5^{x}=4\cdot5^2\cdot5^{29}=4\cdot5^{31}\)
=>\(5^{x}\cdot4=4\cdot5^{31}\)
=>x=31